If $p(x)$ is an $n$ variables polynomial of even degree with complex coefficients which satisfies the strong elliptic condition, that is, Re$p(x) \ge C|x|^{2m}$ for any $x \in \mathbb R^n$ where $2m$ is the degree of $p$. Then can we have the following conclusion:
Foy any $y \neq0 \in \mathbb R^{n-1}$, consider the roots of $p(y,z)=0$ in the complex plane. Then is the number of roots (counting with multiplicity) in the upper half plane equal to that of lower half?
First of all, the notion of ellipticity refers to the principal part of the differential operator. In terms of the symbol, this means a property for the so-called principal symbol. Thus we may assume that $p$ is a homogeneous polynomial, of degree $2m$.
Then the answer to your question is Yes, when $y\in{\mathbb R}^{n-1}$ is non-zero, the univariate polynomial $z\mapsto p(y,z)$ has $m$ roots of positive (resp. negative) imaginary part. Here is the proof. First of all, there are $2m$ roots by assumptions, and none of the roots are real. If $n\ge3$ there imaginary parts keep a constant sign as $y$ varies, because the complement of the origin is connected. Thus let $m_\pm$ be (constant) number of roots of positive/negative imaginary parts. We have $m_-+m_+=2m$. Because the roots $z_j(-y)$ are nothing but the $-z_j(y)$ (by homogeneity), we have $m_-=m_+$. Therefore $m_\pm=m$.
If $n=2$ instead, one build the strongly elliptic polynomial $q(x,t)=p(x)+t^{2m}$ in $3$ variables. The previous analysis applies to $q$ and gives the result for $p$.
Now, the non-homogeneous case. We still know that $z\mapsto p(y,z)$ has $2m$ roots, with a constant number $m_\pm$ of positive/negative imaginary part. Fix $y\ne0$ real and d consider the polynomial $p_s(y,z)=s^{-2m}p(sy,sz)$. When $s\rightarrow+\infty$, $p_s$ tends to the homogeneous part $p_\infty$ of degree $2m$. The assumption implies that $P_\infty$ is strongly elliptic, to which the previous analysis applies. By continuity of the roots as functions of $\frac1s$, we find again that $m_\pm=m$.
